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I am referring to the this wiki page, which states that the way to find the volume of a Polyhedron can be done via Divergence Theorem:

$$\int \nabla \cdot \vec{F} dV=\oint \vec{F} \cdot \hat{n} dS$$

By choosing

$$ \vec{F}=\frac{\vec{x}}{3}$$

One can then obtain the volume formula for a polyhedron with $k$ faces:

$$V = \frac{\sum_{i=1}^k \vec{x_{i}} \cdot \hat{n_i} A_i}{3}$$

where $\vec{x_{i}}$,$\hat{n_i}$ and $A_i$ are a point, normal vector and the area of the face $i$. (The wiki page doesn't say $x_i$ is the centroid; it just says that it can be any point on the face, but I believe that it must be centroid, or else the volume obtained not correct, The $x_i$ can be any point; see comments)

My question is, is my formula and interpretation correct? And does it hold for all kinds of polyhedron, be it convex or concave?

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Green's theorem applies to 2D regions not 3D. Perhaps you meant Divergence theorem? – Aryabhata Jan 2 '11 at 11:10
@Moron, thanks for pointing that out. – Graviton Jan 2 '11 at 11:13
Two points on face $i$ differ by a vector which is orthogonal to $n_i$. It follows that $x_i$ can be any point on face $i$. – Christian Blatter Jan 2 '11 at 12:37
@Christian, now I see it, thanks! Question will be updated accordingly. – Graviton Jan 2 '11 at 12:43
up vote 3 down vote accepted

Yes. The quoted formula holds for any polyhedral surface $S$, or even for a collection of polyhedral surfaces $S_i$, that (together) bound(s) a "volume" $V\subset {\mathbb R}^3$. It is true that a polyhedron has edges and vertices. Therefore, depending on the proof that was originally given for the divergence theorem, one might need an approximation argument to apply the theorem to polyhedra.

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I see. But what about a "cut-through" polyhedron, like the one showing here? – Graviton Jan 3 '11 at 1:45

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